MAT 3105 -Feb. 2007
1. a) Show that the min. poly. m_T(x) of a lin. trans. T is unique and divides any polynomial annihilated by T.
m_T(x) is defined as a polynomial
a_s x^s + … + a_1 x + a_0
such that m_T(T) = 0, a_s = 1, and s is minimum.
Suppose that p(T) = 0, and p has degree d.
We can scale so that p has leading co-efficient 1.
Then, by definition of m_T, s <= d.
The division algorithm for polynomials tells us that there are polynomials q and r such that:
p = mq + r, and 0 <= deg(r) < s.
Now 0 = p(T) = m(T)q(T) + r(T) = 0 q(T) + r(T) = r(T)
So r(T) = 0, and since s is the lowest positive degree of any polynomial such that r(T) = 0, we must have deg(r) = 0.
Thus m divides p exactly.
Moreover, if p is also a min. poly. for T, then:
deg(p) = deg(m) ===> deg(q) = 0, so q is just some constant k.
p and m both have leading co-efficient 1, so k = 1.
Thus p = m, that is, the definition gives us a unique m.
If u is an eigenvalue of T, show that u is a root of m_T(x) = 0.
For the last part, “u is an eigenvector of T” means that Tv = uv, for some non-0 vector v.
Then T^2 v = T(Tv) = T(uv) = u (Tv) = U^2 v, and in general, for all natural numbers n, T^n v = u^n v.
Then m(T)v = a_s T^s v + … + a_1 Tv + a_0 v
= a_s u^s v + … + a_1 u v + a_0 v
= (a_s u^s + … + a_1 u + a_0) v
= m(s) v.
Now m(T) = 0 means that m(T) v = 0,
so m(s) v = 0
and since v is non-0 we have m(s) = 0.
1. c)
If A = (5 -2 // 3 -2) determine the matrix P that
diagonalises A under a similarity operation.
[I use // to denote a newline.]
We want a P such that P’AP is diagonal, where P’ is the inverse of P.
P is the matrix of eigenvectors of A.
Now to find the eigenvectors:
(i) find the eigenvalues using |A – kI| = 0
(ii) solve Av = kv, where k is an eigenvalue
(i) The determinant |A-kI| is:
|5-k -2|
!3 -2-k|
= (5-k)(-2-k) – (-6)
= -[(5-k)(2+k) - 6]
= -[10 +3k - k^2 -6]
= -[k^2 -3k -4]
= -(k-4)(k+1)
So the eigenvalues are k = 4 and k=-1.
(ii) We want to solve Av = kv. Let v = (x // y).
We get:
5x – 2y = kx
3x – 2y = ky
(5-k)x – 2y = 0
3x – (2+k)y = 0
(5-k)x = 2y
3x = (2+k)y
When k=4, we have:
x = 2y
3x = 6y, i.e. x = 2y
so v_4 = (2 // 1)
When k=-1, we have:
6x = 2y, i.e. 3x = y
3x = y
so v_{-1} = (1 // 3)
Thus, one possible P is
(2 1)
(1 3)
with inverse
P’ = 1/5 (3 -1)
(-1 2)
and we can check that P’AP = (4 0)
(0 -1)
[Note: There are at least two ways of getting different matrices P (which will of course mean that P' will be different too).
First, if v is an eigenvector with eigenvalue k, then so is rv, where r is a non-zero scalar. So we could multiply a column of P by any non-0 r.
Second, we could change the order of the columns of P, which in this case will lead to the following diagonal matrix:
(-1 0)
(0 4)
]
Solve A(x_n // y_n) = (x_{n+1} // y_{n+1}) for
(x_0 // y_0) = (3 // -1).
We are given the equation:
A w_n = w_{n+1}, where w_0 is (3 // -1)
To solve the equation, we note that the eigenvectors v_4
and v_{-1} form a basis of R^2.
So we can write w_0 = a v_4 + b v_{-1}
or equivalently w_0 = P (a // b)
so that P’ w_0 = (a // b)
Now P’ w_0 = (2 // -1),
so a = 2 and b = -1, that is:
w_0 = 2 v_4 + (-1) v_{-1}.
Now the equation defines w_n as follows:
w_n = A^n w_0
so w_n = A^n (2 v_4 + (-1) v_{-1})
= 2 A^n v_4 + (-1) A^n v_{-1}
= 2 4^n v_4 + (-1) (-1)^n v_{-1}
= 2 4^n (2 // 1) + (-1)^{n+1} (1 // 3)
= (x_n // y_n)
where x_n = 4^{n+1} + (-1)^{n+1}
and y_n = 2 4^n + 3 (-1)^{n+1}.
